For what values of $x$ is \[\frac{x-10x^2+25x^3}{8-x^3}\]nonnegative?  Answer as an interval.
First we factor $x$ from the numerator, \[\frac{x(1-10x+25x^2)}{8-x^3}.\]Now we see the square of a binomial in the numerator, so our expression is equal to  \[\frac{x(1-5x)^2}{8-x^3}.\]The denominator only has the single (real) root $x=2$, and we can make more sense of that by applying the difference of cubes factorization \[\frac{x(1-5x)^2}{(2-x)(x^2+2x+4)}.\]Now we can factor the entire rational function as  \[\left(\frac{x}{2-x}\right)\left(\frac{(1-5x)^2}{x^2+2x+4}\right).\]Note that the denominator $x^2 + 2x + 4 = (x + 1)^2 + 3$ is always positive.  The factor $x$ changes sign at $x = 0,$ the factor $2 - x$ changes sign at $x = 2,$ and the factor $1 - 5x$ changes sign at $x = \frac{1}{5}.$  We build a sign chart accordingly.

\[
\begin{array}{c|c|c|c|c}
& x < 0 & 0 < x < \frac{1}{5} & \frac{1}{5} < x < 2 & 2 < x \\ \hline
x & - & + & + & + \\
2 - x & + & + & + & - \\
(1 - 5x)^2 & + & + & + & + \\
\left(\frac{x}{2-x}\right)\left(\frac{(1-5x)^2}{x^2+2x+4}\right) & - & + & + & -
\end{array}
\]Also, the expression
\[\left(\frac{x}{2-x}\right)\left(\frac{(1-5x)^2}{x^2+2x+4}\right)\]is equal to 0 at $x = 0$ and $x = \frac{1}{5},$ so the solution to
\[\left(\frac{x}{2-x}\right)\left(\frac{(1-5x)^2}{x^2+2x+4}\right) \ge 0\]is $x \in \boxed{[0,2)}.$